Totally stuck on non exact differential equation: (x - sqrt{x^2 + y^2}) dx + y dy = 0

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Bliman

Junior Member
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Jul 6, 2020
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52
Hi all,
I am totally stuck on an exercise in the differential equation book from Zill. I even have the solution but I just don't understand it. It is in the section about exact differential equations.
It goes like this:


"Differential equations are sometimes solved by having a clever idea. Here is a little exercise in cleverness. Although the differential equation
[imath]\left(x - \sqrt{x^2 + y^2}\right) dx + y\, dy = 0[/imath] is not exact, show how the rearrangement [imath]\frac{x\,dx + y\, dy}{\sqrt{x^2 + y^2}} = dx[/imath] and the observation [imath]\left(\frac{1}{2}\right)\,d(x^2 + y^2) = x\, dx + y\, dy[/imath] can lead to a solution."


I am totally stuck on this exercise. From the first equation, I can rearrange it just fine, but then I am just stuck.
What is even worse is that I have the solution book but I just can't follow it.
It states:


"First note that [imath]d(x^2 + y^2) = \left(\frac{x}{\sqrt{x^2 + y^2}}\right)dx + \left(\frac{y}{\sqrt{x^2 + y^2}}\right)dy[/imath]"


I already have no clue how they come to this.
If I take this [imath]\left(\frac{1}{2}\right)\left(d(x^2 + y^2)\right) = x\, dx + y\, dy[/imath] and multiply both sides with [imath]2[/imath] and divide by [imath](x^2 + y^2)[/imath],
then I will get [imath]d\left(\sqrt{x^2 + y^2}\right) = \left(\frac{2x}{\sqrt{x^2 + y^2}}\right) dx + \left(\frac{2y}{\sqrt{x^2 + y^2}}\right) dy[/imath]
But that is already different than the book.

I am totally stuck on this problem with this problem and I was hoping someone could make it clearer to me.
I also don't really know what this means "[imath]\left(\frac{1}{2}\right)\, d(x^2 + y^2) = x\, dx + y\, dy[/imath]" I don't know what that d means (is that d(function)/dy?) and how they come to that solution.
Like you see I am totally stuck
 
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media-d82-d8247196-5194-46a3-80b9-0b5b635d3836-image.png
Distribute the [imath]dx[/imath]

[imath]x\,dx - \sqrt{x^2+y^2} dx + y\,dy=0[/imath]

Now solve for [imath]x\,dx+y\,dy[/imath]

[imath]x\,dx+y\,dx = \sqrt{x^2+y^2}\,dx[/imath]

Now divide by [imath]\sqrt{x^2+y^2}[/imath]

[imath]\frac{x\,x+y\,dy}{\sqrt{x^2+y^2}} = dx[/imath]

Did that clear things up?
 
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Original differential equation.

\(\displaystyle (x - \sqrt{x^2 + y^2}) \ dx + y \ dy = 0\)

After rearrangement.

\(\displaystyle \frac{(x \ dx + y \ dy)}{\sqrt{x^2 + y^2}} = dx\)

The observation is:

\(\displaystyle \frac{1}{2}d(x^2 + y^2) = x \ dx + y \ dy\)

Then, we have.

\(\displaystyle \frac{\frac{1}{2}d(x^2 + y^2)}{\sqrt{x^2 + y^2}} = dx\)

Let \(\displaystyle z = x^2 + y^2\).

\(\displaystyle \frac{\frac{1}{2}dz}{\sqrt{z}} = dx\)

Now it is easy to solve this differential equation.

\(\displaystyle \frac{1}{2}\int \frac{1}{\sqrt{z}} \ dz = \int dx\)

After solving, undo the substitution \(\displaystyle z = x^2 + y^2\).
 
nasi112 said:
Original differential equation.

\(\displaystyle (x - \sqrt{x^2 + y^2}) \ dx + y \ dy = 0\)

After rearrangement.

\(\displaystyle \frac{(x \ dx + y \ dy)}{\sqrt{x^2 + y^2}} = dx\)

The observation is:

\(\displaystyle \frac{1}{2}d(x^2 + y^2) = x \ dx + y \ dy\)

Then, we have.

\(\displaystyle \frac{\frac{1}{2}d(x^2 + y^2)}{\sqrt{x^2 + y^2}} = dx\)

Let \(\displaystyle z = x^2 + y^2\).

\(\displaystyle \frac{\frac{1}{2}dz}{\sqrt{z}} = dx\)

Now it is easy to solve this differential equation.

\(\displaystyle \frac{1}{2}\int \frac{1}{\sqrt{z}} \ dz = \int dx\)

After solving, undo the substitution \(\displaystyle z = x^2 + y^2\).
Click to expand...
Oh. Thank you so much. It is so easy when it is laid out like that. It is frustrating that I didn't see it.
I would like to learn from this. Because it is so easy when I read it but I can't come up with it by myself.
I would like to ask when you saw my question (next time I will just scan what I have written so it helps with the readability, sorry for that. Just thought about it) how you went and solve it.
Was it clear from the get go or what did you do to come to the answer? Especially the z=x^2+y^2. I would not have come up with this. I am always fearful because I bang my head against such things and my head gets scrambled and I get stuck and can't go further sometimes. I am then afraid I am just not smart enough (I learn all material by myself and with the help of the internet). But I am willing to constantly learn.
Thanks again by the way.
 
Bliman said:
Oh. Thank you so much. It is so easy when it is laid out like that. It is frustrating that I didn't see it.
I would like to learn from this. Because it is so easy when I read it but I can't come up with it by myself.
I would like to ask when you saw my question (next time I will just scan what I have written so it helps with the readability, sorry for that. Just thought about it) how you went and solve it.
Was it clear from the get go or what did you do to come to the answer? Especially the z=x^2+y^2. I would not have come up with this. I am always fearful because I bang my head against such things and my head gets scrambled and I get stuck and can't go further sometimes. I am then afraid I am just not smart enough (I learn all material by myself and with the help of the internet). But I am willing to constantly learn.
Thanks again by the way.
Click to expand...
You're welcome.

When you first posted the problem, it was not Latex organized, so I ignored it. But when you uploaded the picture, I was happy that I could read it clearly. I came up with the idea from the hint ( Observation ). I remember that a few years ago when I was studying complex analysis, some of the problems were easily solved by figuring out the differential element, such as \(\displaystyle d(x^2 + y^2)\).

Without this hint, I am sure that I will be struggling for awhile before answering the problem. It is not a shame if you could not see this simple idea from the first glance. But from now on, this simple trick should reside in your brain hopefully it will help you attack some other differential equations in the future. To be stronger in DE, solve a lot of problems, collect skills and ideas, and don't be disappointed if someday with all these skills, you could not solve a differential equation. It is a very wide topic and cannot be mastered even by the most brightest minds in mathematics.

One last thing, I want to advise you. It is better to learn Latex than scanning problems. It will take you no time to master it and it will help you and other people to read your problems easily and faster.
 
nasi112 said:
You're welcome.

When you first posted the problem, it was not Latex organized, so I ignored it. But when you uploaded the picture, I was happy that I could read it clearly. I came up with the idea from the hint ( Observation ). I remember that a few years ago when I was studying complex analysis, some of the problems were easily solved by figuring out the differential element, such as \(\displaystyle d(x^2 + y^2)\).

Without this hint, I am sure that I will be struggling for awhile before answering the problem. It is not a shame if you could not see this simple idea from the first glance. But from now on, this simple trick should reside in your brain hopefully it will help you attack some other differential equations in the future. To be stronger in DE, solve a lot of problems, collect skills and ideas, and don't be disappointed if someday with all these skills, you could not solve a differential equation. It is a very wide topic and cannot be mastered even by the most brightest minds in mathematics.

One last thing, I want to advise you. It is better to learn Latex than scanning problems. It will take you no time to master it and it will help you and other people to read your problems easily and faster.
Click to expand...
First I want to thank you. I find it very nice and helpful that you answered me.
Yeah when I read the problem I saw it was not readable. I will certainly look at Latex and try to learn it. Thank you for giving insight in the process.
This place is a great place.
 
Bliman said:
I will certainly look at Latex and try to learn it.
Click to expand...
I don't have access to the back end, so I'm not certain, but I *think* that the formatting script used here is KaTeX, which is a derivative of LaTeX. You can insert math formatting by clicking on the "f(x)" button above the message-entry box. "Block" means that the result will be centered on its own line; "Inline" means that the result will be included in the flow of your text.

You can also insert math formatting yourself, by using tags. This looks like the following:

Code: 
I am writing math within a sentence, as [imath]ax^2 + bx + c = 0[/imath]. Then I'll put some math centered on its own line:

[math]x = \dfrac{-b \pm \sqrt{b^2 - 4ac\,}}{2a}[/math]

See how nice that looks?

The above displays as follows:


I am writing math within a sentence, as [imath]ax^2 + bx + c = 0[/imath]. Then I'll put some math centered on its own line:

[math]x = \dfrac{-b \pm \sqrt{b^2 - 4ac\,}}{2a}[/math]
See how nice that looks?


You can see how KaTeX formats stuff on this page.

Eliz.
 
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